Using Appendix F, along with the fact that the earth spins on its axis once per day, calculate (a) the earth’s orbital angular speed (in rad/s) due to its motion around the sun, (b) its angular speed (in rad/s) due to its axial spin, (c) the tangential speed of the earth around the sun (assuming a circular orbit), (d) the tangential speed of a point on the earth’s equator due to the planet’s axial spin, and (e) the radial and tangential acceleration components of the point in part (d).

Solution 19E Step 1: (a).As the earth orbits the sun,the earth makes 1 revolution(2 rad)in one year.taking the positive direction for the angular displacement to be the direction of the earth’s orbital motion,the angular displacement in one year is = 2.the average angular velocity is orbit 365 ¼ days convert into seconds. w = () orbit(t)orbit w = 2 rad/(365 1/4 days)(24 h/1 day)(3600 s/1 h) w = 2.0 × 10 7 rad/s. Step 2: (b).As the earth spins on its axis,it makes one revolution (2 rad) in a day.assuming that the positive direction for the angular displacement is the same as direction of the earth’s rotation, the angular displacement of the earth in one day is spin= 2 rad.the average angular velocity is w = () spin(t)spin w = 2 rad/(1 day)(24 h/1 day)(3600 s/1 h) w = 7.3 × 10 5 rad/s Step 3: (C).The tangential speed of the earth around the sun V = earth orbital angular speed × radius of earth orbit tan V = (1.5 × 10 m) × (2.0 × 10 7rad/s) tan 4 V tan= 3 × 10 m/s